Problem: Let $g(x)=\dfrac{x^2-3}{x+2}$ Where does $g$ have critical points? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=-3$ (Choice B) B $x=-2$ (Choice C) C $x=-1$ (Choice D) D $g$ has no critical points.
A critical point of $g$ is a point in the domain of $g$ where the derivative is either equal to zero or undefined. So in order to find the critical points of $g$, let's find its derivative. $\begin{aligned} g'(x)&=\dfrac{d}{dx}\left[ \dfrac{x^2-3}{x+2} \right] \\\\ &=\dfrac{\dfrac{d}{dx}[x^2-3](x+2)-(x^2-3)\dfrac{d}{dx}[x+2]}{(x+2)^2} \\\\ &=\dfrac{(2x)(x+2)-(x^2-3)(1)}{(x+2)^2} \\\\ &=\dfrac{x^2+4x+3}{(x+2)^2} \\\\ &=\dfrac{(x+1)(x+3)}{(x+2)^2} \end{aligned}$ Now let's look for $x$ -values where $g'$ is zero or undefined. $\dfrac{(x+1)(x+3)}{(x+2)^2}=0$ at $x=-3$ and $x=-1$. $\dfrac{(x+1)(x+3)}{(x+2)^2}$ is undefined at $x=-2$. However, $g(x)=\dfrac{x^2-3}{x+2}$ is also undefined at $x=-2$ so it isn't a critical point. In conclusion, these are the $x$ -values where $g$ has critical points: $x=-3$ $x=-1$